The equation of the line passing through the point `(1,2)` and perpendicular to the line `x+y+1=0` is

Options

A. `y-x+1=0`
B. `y-x-1=0`
C. `y-x+2=0`
D. `y-x-2=0`

Explanation

To find the equation of the line passing through the point (1, 2) and perpendicular to the line given by the equation \(x + y + 1 = 0\), we can follow these steps:

### Step 1: Find the slope of the given line
The equation of the line is in the form \(Ax + By + C = 0\). Here, \(A = 1\), \(B = 1\), and \(C = 1\).

The slope \(m\) of a line in this form is given by:
\[ m = -\frac{A}{B} \]
Substituting the values:
\[ m = -\frac{1}{1} = -1 \]

### Step 2: Find the slope of the perpendicular line
If two lines are perpendicular, the product of their slopes is \(-1\). Let the slope of the perpendicular line be \(m'\). Therefore:
\[ m \cdot m' = -1 \]
Substituting \(m = -1\):
\[ -1 \cdot m' = -1 \]
From this, we find:
\[ m' = 1 \]

### Step 3: Use the point-slope form to find the equation of the line
The point-slope form of the equation of a line is given by:
\[ y - y_1 = m'(x - x_1) \]
Here, \((x_1, y_1) = (1, 2)\) and \(m' = 1\). Substituting these values into the equation:
\[ y - 2 = 1(x - 1) \]

### Step 4: Simplify the equation
Expanding the equation:
\[ y - 2 = x - 1 \]
Rearranging gives:
\[ y - x + 1 = 0 \]
Or equivalently:
\[ y - x - 1 = 0 \]

### Final Answer
The equation of the line passing through the point (1, 2) and perpendicular to the line \(x + y + 1 = 0\) is:
\[ y - x - 1 = 0 \]